Oracle 查询Long类型

好久以前就碰过这个问题，今天找出原来写的一个函数

select b.remark,
       LONG_TO_CHAR(b.rowid, null, 'tmp20180518102050381', 'remark')
  from tmp20180518102050381 b;

 

 

CREATE OR REPLACE fUNCTION LONG_TO_CHAR(in_rowid      rowid, --rowid
                                        in_owner      varchar, --所有者   可空
                                        in_table_name varchar, --表名
                                        in_column     varchar2 --要查询的列名
                                        ) RETURN varchar AS
  text_c1 varchar2(32767);
  sql_cur varchar2(2000);
begin
  if length(in_owner) > 0 then
    sql_cur := 'select ' || in_column || ' from ' || in_owner || '.' ||
               in_table_name || ' where rowid = ' || chr(39) || in_rowid ||
               chr(39);
  else
    sql_cur := 'select ' || in_column || ' from ' || in_table_name ||
               ' where rowid =' || chr(39) || in_rowid || chr(39);
  end if;
  dbms_output.put_line(sql_cur);
  execute immediate sql_cur
    into text_c1;

  text_c1 := substr(text_c1, 1, 4000);
  RETURN TEXT_C1;
END;

 

来源：https://www.cnblogs.com/power255/p/9176974.html